Question

One mole of an ideal monoatomic gas undergoes the process $P=\alpha T^{1/2}$, where $\alpha$ is a constant. Find the work done by the gas if its temperature increases by $50K$.

• A. $407.75J$
• B. $207.75J$
• C. $307.75J$
• D. $107.75J$

Answer 1

Answer: (B)

$207.75J$

The given process is $P=\alpha T^{1/2}$

$⟹$ $P{T}^{-1/2}=\alpha =constant$

Comparing with $P{T}^{\frac{m}{1-m}}=constant$ we get $\frac{m}{1-m}=\frac{-1}{2}$

$⟹$ $m=-1$

For monoatomic gas $\gamma =1.67$

Molar specific heat of the gas $C=\frac{R}{\gamma -1}+\frac{R}{1-m}$

$\therefore$ $C=\frac{R}{1.67-1}+\frac{R}{1-(-1)}=2R$

Molar specific heat of the gas $C=2R$

Given : $\Delta T=50K$

Heat supplied to the gas $Q=nC\Delta T$ where $n=1$

$\therefore$ $Q=2R\times 50=100R$

Change in internal energy $\Delta U=\frac{nR\Delta T}{\gamma -1}$

$\therefore$ $\Delta U=\frac{1R\times 50}{1.67-1}=75R$

From 1st law $Q=\Delta U+W$

$\therefore$ $100R=75R+W$

$⟹W=25R=25\times 8.314=207.75$ J